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**Unformatted text preview: **Math 16b Thomas Scanlon Autumn 2010 Thomas Scanlon Math 16b Chain rule recalled (f · g ) (x ) = f (x )g (x ) + f (x )g (x ) Thomas Scanlon Math 16b Inverting the Chain Rule: Integration by Substitution (f · g ) (x ) = f (x )g (x ) + f (x )g (x ) f (x )g (x ) = (f · g ) (x ) − f (x )g (x ) f (x )g (x )dx =
(f · g ) (x )dx − g (x )f (x )dx g (x )f (x )dx f (x )g (x )dx = f (x )g (x ) − Thomas Scanlon Math 16b Formalism of Integration by Parts Often, one nds two functions u and v so that the integrand may be written as udv where dv = v (x )dx . If we succeed in so doing, then from the equality u (x )v (x ) = u (x )v (x ) − v (x )u (x )
we see that if f (x ) = u (x )v (x ), then f (x )dx = u (x )v (x )dx = udv = uv − vdu If vdu is easier to evaluate than is succeeds.
Thomas Scanlon Math 16b f (x )dx , then the method An integral revisited Take u = x and v = − cos(x ) so that dv = sin(x )dx and du = dx . Then, x sin(x )dx = = u dv uv − v du
cos(x )dx = −x cos(x ) + = −x cos(x ) + sin(x ) + C Thomas Scanlon Math 16b Example Integrate: xe x dx
Take u = x and v = e x so that du = dx and dv = e x dx . x e x dx = = = = u dv uv − xe x − v du e x dx xe x − e x + C Thomas Scanlon Math 16b Another example Integrate e x cos(x )dx
Set u = e x and v = sin(x ) so that du = e x dx and dv = cos(x )dx . Then e x cos(x )dx = = = = u dv uv − v du e x sin(x )dx e x sin(x )dx e x sin(x ) − e x sin(x ) − Thomas Scanlon Math 16b Solution, continued e x cos(x )dx = e x sin(x ) − e x sin(x )dx e x cos(x )dx e x cos(x )dx e x cos(x )dx = e x sin(x ) + e x cos(x ) − e x cos(x )dx = e x sin(x ) + e x cos(x ) −
2 e x cos(x )dx = e x sin(x ) + e x cos(x ) Thomas Scanlon Math 16b 1 1 2 2 x and y = cos(x ) so that dw = e x dx and Set w = e dy = − sin(x )dx Then
x e x cos(x )dx = e x sin(x ) + e x cos(x ) + C A third example Integrate ln(x )dx Set u = ln(x ) and v = x so that du = ln(x )dx =
= = = =
Thomas Scanlon Math 16b dx x and dv = dx . u dv uv − v du x dx x dx
1 x ln(x ) − x ln(x ) − x ln(x ) − x + C ...

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